2009 AMC 10B Problem 18

Below is the professionally curated solution for Problem 18 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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Difficulty rating: 1370

18.

Rectangle $ABCD$ has $AB=8$ and $BC=6.$ Point $M$ is the midpoint of diagonal $\overline{AC},$ and $E$ is on $\overline{AB}$ with $\overline{ME}\perp\overline{AC}.$ What is the area of $\triangle AME?$

$\dfrac{65}{8}$

$\dfrac{25}{3}$

$9$

$\dfrac{75}{8}$

$\dfrac{85}{8}$

Solution:

By the Pythagorean Theorem, $AC=\sqrt{8^2+6^2}=10,$ so $AM=5.$ Right triangles $AME$ and $ABC$ share angle $A,$ so they are similar with $\dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68,$ giving $ME=\dfrac{15}{4}.$

Then $\text{area}(\triangle AME)=\dfrac12\cdot AM\cdot ME=\dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8}.$

Thus, the correct answer is D.

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